Simplification of hadamard product $K^{-1}((Ka1^T)\circ I)$ where $K=X^TX$

Question

I think $K^{-1}((Ka1^T)\circ I)$ could be simplified in some way. Here, $I$ is an identity matrix. And K could be expressed as $X^TX$ for some $X$. $\circ$ is hadamard product. $a$ is a vector. $1^T$ is a row vector where all elements are one.

Is there a way to eliminate $K^{-1}$?

You could add some assumptions if needed.

Answer 1

If you multiply the whole expression by $\def\o{{\tt1}}\o$ then $$\eqalign{ \def\D{\operatorname{Diag}} \def\K{K^{-1}} \K\Big((Ka\o^T)\odot I\Big)\,\o &= \K\Big(\D(Ka)\Big)\,\o \\ &= \K\big(Ka\big) \\&= a \\ }$$ Otherwise you get stuck after the first step $$\eqalign{ \K\Big((Ka\o^T)\odot I\Big) &= \K\,\D(Ka) \\ }$$